If $(P,\leq)$ is a pre-odered set (that is, $\leq$ is a reflexive and transitive relation) and $x\in P$, we set $(\uparrow_{\leq} x) = \{p\in P: p\geq x\}$ and $(\downarrow_{\leq} x) = \{p\in P: p\leq x\}$.

Let $\text{NPU}(\omega)$ be the set of non-principal ultafilters on $\omega$. The Rudin-Keisler preorder on $\text{NPU}(\omega)$ is defined by $${\cal U} \leq_{RK} {\cal V} :\Leftrightarrow (\exists f:\omega\to\omega)(\forall U\in{\cal U}) f^{-1}(U)\in {\cal V} .$$

The Wallman topology on $\text{NPU}(\omega)$ is defined as follows: For $A\subseteq \omega$ set $\Phi_A = \{{\cal U} \in \text{NPU}(\omega): A\in{\cal U}\}$. Then $${\frak S} = \big\{\text{NPU}(\omega) \setminus \Phi_A: A\subseteq \omega\big\}$$ is a subbasis for the Wallman topology on $\text{NPU}(\omega)$.

Question. Pick any ${\cal U}\in \text{NPU}(\omega)$. Are $(\downarrow_{\leq_{RK}} {\cal U})$ and $(\uparrow_{\leq_{RK}} {\cal U})$ closed in the Wallman topology on $\text{NPU}(\omega)$?

(Note. This would imply that the Wallman topology contains the interval topology of $(\text{NPU}(\omega), \leq_{RK})$.)


1 Answer 1


As far as I can see, the Wallman topology as defined here is the same as the topology that $NPU(\omega)$ gets as a subspace of the Stone-Cech compactification of the discrete space $\omega$. Specifically, for any $A\subseteq\omega$, any ultrafilter on $\omega$ contains either $A$ or $\omega-A$ but not both, so the subbasic open set $NPU(\omega)-\Phi_A$ equals $\Phi_{\omega-A}$, so your subbasis consists of all the sets of the form $\Phi_A$. Furthermore, since an ultrafilter contains the intersection of two sets iff it contains both of those sets, we have $\Phi_{A\cap B}=\Phi_A\cap\Phi_B$, and so the subbasis is in fact a basis. Since $\Phi_A$ is just the intersection of $NPU(\omega)$ and a typical basic open set $\{\mathcal U\in\beta\omega:A\in\mathcal U\}$, it follows that the topology on $NPU(\omega)$ is just the subspace topology inherited from $\beta\omega$.

Now to address your questions: Neither of the sets that you asked about is ever closed. In fact, for every $\mathcal U\in NPU(\omega)$, its RK-equivalence class is dense in $NPU(\omega)$. To see this, consider any such $\mathcal U$ and any nonempty basic open set $\Phi_A$. Here $A$ is an infinite subset of $\omega$ because if it were finite then $\Phi_A$ would be empty. I intend to find an ultrafilter RK-equivalent to $\mathcal U$ that lies in $\Phi_A$, i.e., that contains $A$. By shrinking $A$ if necessary, we can assume that not only $A$ but also $\omega-A$ is infinite. Furthermore, $\mathcal U$ contains some infinite subset $B$ of $\omega$ whose complement is also infinite; indeed, we can take $B$ to be the set of even numbers or the set of odd numbers, whichever is in $\mathcal U$. (Exactly one of them is in $\mathcal U$ because $\mathcal U$ is an ultraiflter.) So there is a permutation $f$ of $\omega$ mapping $B$ onto $A$. Let $\mathcal V=\{X\subseteq\omega:f^{-1}(X)\in\mathcal U\}$. Then it is easy to check that $\mathcal V$ and $\mathcal U$ are each RK-$\leq$ the other, via the functions $f$ and $f^{-1}$. So $\mathcal V$ is RK-equivalent to $\mathcal U$. And $\mathcal V\in\Phi_A$ because $f^{-1}(A)=B\in\mathcal U$.


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